{"id":6120,"date":"2023-01-21T21:10:04","date_gmt":"2023-01-21T20:10:04","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=6120"},"modified":"2023-01-25T10:25:03","modified_gmt":"2023-01-25T09:25:03","slug":"the-space-s0","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=6120","title":{"rendered":"The space S0"},"content":{"rendered":"\n

The space S<\/em>0<\/sub> originates from a deep paper of Matthew de Brecht’s on a characterization of quasi-Polish spaces by forbidden substructures [1]. Namely, his theorem (Theorem 7.2 in [1]) says that a coanalytic subspace of a quasi-Polish space will always be quasi-Polish unless it embeds one of four possible forbidden substructures as \u03a0<\/strong>2<\/sup>0<\/sub>-subspaces. I will not explain what this all means, except that the four possible substructures in question, which de Brecht calls S<\/em>0<\/sub>, S<\/em>1<\/sub>, S<\/em>2<\/sub> and S<\/em>D<\/em><\/sub>, are well-known, except for the first one. S<\/em>1<\/sub> is the T1<\/sub> space of natural numbers with the cofinite topology, S<\/em>2<\/sub> is the T2<\/sub> space of rational numbers with its usual, metric topology, S<\/em>D<\/em><\/sub> is the TD<\/em><\/sub> space of natural numbers with the Alexandroff (or Scott) topology of its usual ordering. S<\/em>0<\/sub> is a T0<\/sub> space, as you may have guessed, but does not seem to have appeared earlier in the literature.<\/p>\n\n\n\n

A few years ago, Matthew de Brecht was visiting my university, and tried to get me interested in S<\/em>0<\/sub>, and in particular to try to discover whether S<\/em>0<\/sub> was consonant or not. I had a hard time understanding the properties of S<\/em>0<\/sub>, and I was unable to contribute. I was sure he managed to show that S<\/em>0<\/sub> is not consonant, but the paper where I expected to find it [2] makes no mention of consonance at all. If I can sort this out, I will try to explain the answer to the question on this blog. In any case, now I understand S<\/em>0<\/sub> a bit better, and I would like to explain what I know about it.<\/p>\n\n\n\n

The definition of S<\/em>0<\/sub><\/h2>\n\n\n\n

Right, so, here is the definition (not quite the same as in [1], which uses prefixes instead of suffixes, but the space we will obtain will be homeomorphic). S<\/em>0<\/sub> is the space of finite sequences of natural numbers, with the upper topology of the (opposite of the) suffix ordering.<\/p>\n\n\n\n

Let me explain this more slowly. I will write \u03b5 for the empty sequence, n<\/em>::s<\/em> for the sequence obtained by adding a natural number n<\/em> in front of a sequence s<\/em>, and n<\/em>1<\/sub>…n<\/em>k<\/em><\/sub> for the sequence of natural numbers n<\/em>1<\/sub>, …, n<\/em>k<\/em><\/sub>. In particular, n<\/em>1<\/sub>…n<\/em>k<\/em><\/sub> = n<\/em>1<\/sub>::(n<\/em>2<\/sub>…n<\/em>k<\/em><\/sub>) if k<\/em>\u22651, and in general n<\/em>1<\/sub>…n<\/em>k<\/em><\/sub> = n<\/em>1<\/sub>::(n<\/em>2<\/sub>:: … (n<\/em>k<\/em><\/sub>::\u03b5)…).<\/p>\n\n\n\n

A suffix<\/em> of a finite sequence n<\/em>1<\/sub>…n<\/em>k<\/em><\/sub> is any finite sequence n<\/em>p<\/em><\/sub>…n<\/em>k<\/em><\/sub>, where 1\u2264p<\/em>\u2264k<\/em>+1 (if p<\/em>=k<\/em>+1, this denotes the empty sequence). Let me write s<\/em>\u2264t<\/em> if t<\/em> is a suffix of s<\/em>. This is the (opposite of the) suffix ordering. Note that S<\/em>0<\/sub> has a largest element, which is the empty sequence \u03b5. Each element s<\/em> has a countably infinite set of successors<\/em> n<\/em>::s<\/em>, n<\/em> \u2208 N<\/strong>, where is a successor of s<\/em> is an element immediately below s<\/em> (i.e., strictly below s<\/em> and without any other element inbetween). We then see that S<\/em>0<\/sub> is a reversed tree<\/em> (some vertices and some branches are missing \u2014 it’s pretty hard to draw an infinite structure as a finite picture), with root \u03b5:<\/p>\n\n\n

\n
\"\"<\/a>
The reversed tree S0<\/sub><\/figcaption><\/figure>\n<\/div>\n\n\n

We equip S<\/em>0<\/sub> with the upper topology of its ordering \u2264. This means that the closed subsets of S<\/em>0<\/sub> are the intersections \u2229i<\/em> \u2208 I<\/em><\/sub> \u2193Ei<\/sub><\/em>, where each Ei<\/sub><\/em> is a finite set. (\u2193Ei<\/sub><\/em> is the downward closure of Ei<\/sub><\/em>, hence the set of all sequences that have a suffix in Ei<\/sub><\/em>\u2014although one has a better grasp of what they are by imagining what such sets would look like on the picture above. The downwards closed subsets of finite sets are called finitary closed<\/em> subsets in the <\/a>book<\/a>.)<\/p>\n\n\n\n

Since there are countable many finitary closed subsets \u2193E<\/em> in S<\/em>0<\/sub>, let me note in passing that S<\/em>0<\/sub> is not just countable, but also countably-based.<\/p>\n\n\n\n

The closed subsets of S<\/em>0<\/sub><\/h2>\n\n\n\n

Since S<\/em>0<\/sub> is countable, it has only countably many finitary closed subsets. However, as M. de Brecht notices, it has more. That is, it has closed sets that are non-trivial infinite intersections \u2229i<\/em> \u2208 I<\/em><\/sub> \u2193Ei<\/sub><\/em> of finitary closed sets. In fact, it has so many that they form an uncountable<\/em> set:<\/p>\n\n\n\n

Proposition [1, Proposition 6.1].<\/strong> S<\/em>0<\/sub> has uncountably many closed sets. Equivalently, S<\/em>0<\/sub> has uncountably many open sets.<\/p>\n\n\n\n

Proof.<\/em> It suffices to exhibit an injection from the set NN<\/sup><\/strong> of infinite sequences of natural numbers to the collection H<\/strong>(S<\/em>0<\/sub>) of closed subsets of S<\/em>0<\/sub>.<\/p>\n\n\n\n

Given any infinite sequence k<\/strong><\/em>\u225d(kn<\/sub><\/em>)n<\/em> \u2208 N<\/strong><\/sub> of natural numbers, we form the following points of S<\/em>0<\/sub>: first the one-element sequence k<\/em>0<\/sub>+1 (namely, the k<\/em>0<\/sub>+2nd successor of the root \u03b5), then the two-element sequence (k<\/em>1<\/sub>+1)0 (the k<\/em>1<\/sub>+2nd successor of the element 0), then the three-element sequence (k<\/em>2<\/sub>+1)00 (the k<\/em>2<\/sub>+2nd successor of the element 00), and so on. In other words, we look at the leftmost branch of the tree 00…0…, and we pick the k<\/em>0<\/sub>+2nd successor of the root, the k<\/em>1<\/sub>+2nd successor of the second element from the root on that branch, etc. In general, element number n<\/em> of that sequence of points is uk<\/sup><\/strong>n<\/sub><\/em> \u225d (k<\/em>n<\/em><\/sub>+1)0n<\/sup><\/em>, where 0n<\/sup><\/em> denotes a sequence of n<\/em> zeroes. (The superscript k<\/strong><\/em> marks the dependence on the initial infinite sequence k<\/em><\/strong>.)<\/p>\n\n\n\n

We define Fk<\/sup><\/strong><\/em><\/sub><\/em> as the downward closure of the infinite set {uk<\/sup><\/strong>n<\/sub><\/em> | n<\/em> \u2208 N<\/strong>}. For example, here is Fk<\/sup><\/strong><\/em><\/sub><\/em> when k<\/em><\/strong> is the sequence of natural numbers 2, 1, 0, … (again, due to space limitations, I am only showing the downward closures of the first three points uk<\/sup><\/strong><\/em>0<\/sub>, uk<\/sup><\/strong><\/em>1<\/sub>, and uk<\/sup><\/strong><\/em>2<\/sub>). I have shown it in blue:<\/p>\n\n\n

\n
\"\"<\/a><\/figure>\n<\/div>\n\n\n

I claim that Fk<\/sup><\/strong><\/em><\/sub><\/em> is closed in S<\/em>0<\/sub>. To this end, for each natural number n<\/em>, let Ek<\/sup><\/strong><\/em>n<\/sub><\/em> be the finite set consisting of u<\/em>k<\/sup><\/strong><\/em>0<\/sub>, …, uk<\/sup><\/strong><\/em>n<\/sub><\/em>, plus the point 0n<\/em>+1<\/sup>. On the picture above, and taking n<\/em>\u225d2, that would be the collection of the top elements u<\/em>k<\/sup><\/strong><\/em>0<\/sub>, uk<\/sup><\/strong><\/em><\/sub><\/em>1<\/sub>, uk<\/sup><\/strong><\/em><\/sub><\/em>2<\/sub> of each of the displayed blue regions, plus the point 000 at the bottom left of the picture. It is pretty easy to see that Fk<\/sup><\/strong><\/em><\/sub><\/em> \u225d \u2229n<\/em> \u2208 N<\/strong><\/sub> \u2193Ek<\/sup><\/strong><\/em><\/sub><\/em>n<\/sub><\/em>, and therefore Fk<\/sup><\/strong><\/em><\/sub><\/em> is closed.<\/p>\n\n\n\n

It is also easy to see that the map k<\/em><\/strong> \u21a6 Fk<\/sup><\/strong><\/em><\/sub><\/em> is injective. Let us imagine another infinite sequence k<\/strong><\/em>‘<\/em><\/strong>\u225d(k’n<\/sub><\/em>)n<\/em> \u2208 N<\/strong><\/sub> of natural numbers. Since it is distinct from k<\/em><\/strong>, we have kn<\/sub><\/em>\u2260k’n<\/sub><\/em> for some index n<\/em>. The points of Fk<\/sup><\/strong><\/em><\/sub><\/em> of the form …k<\/em>0n<\/sup><\/em> with k<\/em>\u22600 are exactly those such that k<\/em>=kn<\/sub><\/em>+1, and similarly, those of Fk’<\/sup><\/strong><\/em><\/sub><\/em> of the form …k<\/em>0n<\/sup><\/em> with k<\/em>\u22600 are those such that k<\/em>=k’n<\/sub><\/em>+1. Hence, for example, (kn<\/sub><\/em>+1)0n<\/sup><\/em> is in Fk<\/sup><\/strong><\/em><\/sub><\/em> but not in Fk’<\/sup><\/strong><\/em><\/sub><\/em>. (And (k’n<\/sub><\/em>+1)0n<\/sup><\/em> is in Fk’<\/sup><\/strong><\/em><\/sub><\/em> but not in Fk<\/sup><\/strong><\/em><\/sub><\/em>.) \u2610<\/p>\n\n\n\n

So S<\/em>0<\/sub> is a countable space, even a second-countable space, with an uncountable topology. That is not a big deal. The same could be said of the space Q<\/strong> of rational numbers, with its usual topology. M. de Brecht also shows that S<\/em>0<\/sub> is not a TD<\/sub> space, and I will leave that as an exercise to you. (First read the characterization of closed sets, which will come below.)<\/p>\n\n\n\n

Sobriety<\/h2>\n\n\n\n

We now have:<\/p>\n\n\n\n

Proposition.<\/strong> S<\/em>0<\/sub> is sober.<\/p>\n\n\n\n

This follows from a pretty useful theorem from Andrea Schalk’s PhD thesis:<\/p>\n\n\n\n

Proposition [3, Proposition 1.7].<\/strong> Every poset P<\/em> in which every non-empty family has a supremum is sober in its upper topology.<\/p>\n\n\n\n

(I have already cited this proposition from Schalk in this post<\/a>, but not with this level of generality.)<\/p>\n\n\n\n

Proof.<\/em> Since P<\/em> is a poset, P<\/em> is T0<\/sub> in its upper topology.<\/p>\n\n\n\n

Let C<\/em> be an irreducible closed subset of P<\/em>. Since C<\/em> is non-empty, it has a supremum x<\/em>, by assumption. We claim that x<\/em> is in C<\/em>. Otherwise, the complement P<\/em>\u2013C<\/em> of C<\/em> would be an open neighborhood of x<\/em>, and would therefore contain a basic open set P<\/em>\u2013\u2193E<\/em> containing x<\/em>, where E<\/em> is finite. Since P<\/em>\u2013C<\/em> contains P<\/em>\u2013\u2193E<\/em>, we must have C<\/em> \u2286 \u2193E<\/em>. But C<\/em> is irreducible and \u2193E<\/em> is the finite union of sets \u2193y<\/em>, y<\/em> \u2208 E<\/em>, so C<\/em> is included in \u2193y<\/em> for some y<\/em> \u2208 E<\/em>. That implies that y<\/em> is an upper bound of C<\/em>. Since x<\/em> is the least one, x<\/em>\u2264y<\/em>; but then x<\/em> is in \u2193E<\/em> as well, contradicting the fact that x<\/em> is in P<\/em>\u2013\u2193E<\/em>.<\/p>\n\n\n\n

Therefore x<\/em> is the largest element of C<\/em>, and hence C<\/em> = \u2193x<\/em>. \u2610<\/p>\n\n\n\n

We apply this to S<\/em>0<\/sub>: given any family F<\/em> of points of S<\/em>0<\/sub>, namely of finite sequences of natural numbers, it should be fairly clear that all of them have a longest common suffix; that suffix is the supremum of F<\/em>. Then the latter proposition immediately gives us that S<\/em>0<\/sub> is sober.<\/p>\n\n\n\n

Being sober, S<\/em>0<\/sub> is a monotone convergence space (Proposition 8.2.34 in the <\/a>book<\/a>). This means that:<\/p>\n\n\n\n